Question

Which of the following methods could you use to prepare 1.00 L of 0.225 M HI?

Place 28.8 g of HI in a 1.00-L volumetric flask and then add enough water to make 1.00 L of solution

.Measure out 37.5 mL of 6.00 M HI and dilute with enough water to prepare 1.00 L of solution.

Dilute 75.0 mL of 3.00 M HI with 925.0 ml of water.

Combine 28.8 g of HI with 1.00 L water.

Mix 37.5 mL of 6.00 M HI and 962.5 ml of water.

Answer 1

Place 28.8 g of HI in a 1.00-L volumetric flask and then add enough water to make 1.00 L of solution

MW of HI = 127.904

mass needed for 0.225 mol

mass = mol*MW = 0.225*127.904 = 28.8 g of HI....

Therefore

the first option is true, if oyu mix 28.8 g of HI with enough water to hav ea total volume of 1 Liter... you will get a 0.225 mix

Measure out 37.5 mL of 6.00 M HI and dilute with enough water to prepare 1.00 L of solution.

moles of HI present in total = M*V = 6*(37.5/1000) = 0.225 moles of HI

If you mix with enough water... then M = 0.225 mols per liter... this is also TRUE

Dilute 75.0 mL of 3.00 M HI with 925.0 ml of water.

Total amount found of moles:

M*v = 3*75/1000 = 0.225 mol of HI

925 ml + 75 ml = 1000 ml or 1 liter

therefore, 0.225 mol in 1 liter, this is also TRUE .. NOTE that in real life, you cannot assume volumes amay be added, that is 75 ml of a mixture plus 975 ml of another substance will not always give you 1000 ml... in this case it may be considered ok

Combine 28.8 g of HI with 1.00 L water.

PPretty similar as case I... we have 0.225 mol in 1 liter, this is also TRUE ..

NOTE that in real life, you cannot assume volumes amay be added, that is 75 ml of a mixture plus 975 ml of another substance will not always give you 1000 ml... in this case it may be considered ok

Mix 37.5 mL of 6.00 M HI and 962.5 ml of water.

This will also get you 0.225 mol of HI, the total volume, if you consider addition of volume constans, will give you 1 Liter...

FINAL NOTE

IF volumes may not be added together, only solution 1 and 2 are true